Simplify; express your answer in exponential form. Assume $k\neq 0, z\neq 0$. $\dfrac{{(k^{3}z^{-5})^{5}}}{{(k^{3}z^{-1})^{5}}}$
Solution: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(k^{3}z^{-5})^{5} = (k^{3})^{5}(z^{-5})^{5}}$ On the left, we have ${k^{3}}$ to the exponent ${5}$ . Now ${3 \times 5 = 15}$ , so ${(k^{3})^{5} = k^{15}}$ Apply the ideas above to simplify the equation. $\dfrac{{(k^{3}z^{-5})^{5}}}{{(k^{3}z^{-1})^{5}}} = \dfrac{{k^{15}z^{-25}}}{{k^{15}z^{-5}}}$ Break up the equation by variable and simplify. $\dfrac{{k^{15}z^{-25}}}{{k^{15}z^{-5}}} = \dfrac{{k^{15}}}{{k^{15}}} \cdot \dfrac{{z^{-25}}}{{z^{-5}}} = k^{{15} - {15}} \cdot z^{{-25} - {(-5)}} = z^{-20}$